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Joule heating is the dissipative conversion of electrical work into internal energy in an electrical conductor. In a metal, the electric field supplies energy to charge carriers, which transfer it to the material. This makes an immersed resistor a useful calibrated heater and makes electrical resistance a source of losses in wiring. Wind-driven electric currents are a proposed source of heating of an inflated giant planet.

For a stationary, purely resistive Ohmic conductor, use the passive sign convention: $V$ is the terminal electric potential difference from the entry to the exit of positive electric current $I$. The absorbed electrical power and Ohm's law giveThe last two expressions assume $V=IR$ at the relevant material temperature. For constant electrical resistance and electric current over a duration $\Delta t$, the dissipated energy is $E_{\mathrm{diss}}=I^{2}R\Delta t$; otherwise use $E_{\mathrm{diss}}=\int V(t)I(t)dt$. In SI, $\mathrm{V}\mathrm{A}=\mathrm{W}=J/s$.^[ref]

Absorbed electrical power $VI$ is more general than Joule heating: a capacitor can store energy, a battery can change its chemical energy, and an electrical circuit can deliver mechanical work or emit light. Even a heated resistor need not show an indefinitely rising temperature: in a thermal steady state, its dissipated power leaves by heat transfer. With the heater inside a calorimeter boundary, electrical input crosses as work; the subsequent heat transfer to the room is a different term in the first law of thermodynamics.

The passive sign convention defines terminal electric potential difference $V=\varphi (A)-\varphi (B)$ and positive electric current entering at $A$. Then $VI$ is absorbed electrical power: positive means the component receives energy, negative means it supplies energy. Reversing both reference directions leaves the product unchanged. For a positive electrical resistance obeying Ohm's law, both $V$ and $I$ reverse together and Joule heating remains $I^{2}R\ge 0$.

With the passive sign convention applied to the network, its entering electric current is $J=-I$, and the same relation becomes $V=V_{\mathrm{oc}}+J{R}_{\mathrm{th}}$. The electrical load uses the passive sign convention with entering electric current $I$, so it absorbs power $VI$. A positive delivered electric current therefore produces a voltage drop for positive $R_{\mathrm{th}}$; changing the reference direction without changing the equation would give the wrong sign.

The reason is linearity of Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and the constitutive relations of any linear dependent electrical sources. Solve once at $I=0$ with the independent electrical sources present. Subtract this solution from a solution with prescribed delivered electric current $I$: the fixed independent electrical sources cancel, leaving a homogeneous linear network whose terminal voltage change is proportional to $I$. This is a short superposition principle argument for an affine function; the fixed offset need not vanish.

The qualification about a well-defined response matters. An ideal voltage source alone has $R_{\mathrm{th}}=0$. An ideal nonzero current source alone constrains $I$ and does not have a finite open-circuit voltage or a representation $V=V_{\mathrm{oc}}-I{R}_{\mathrm{th}}$ with finite coefficients. Linear dependent electrical sources can give negative resistance, and singular ideal networks may fail to have a unique finite response. Neither a finite nor a strictly positive passive $R_{\mathrm{th}}$ follows merely from calling an electrical circuit linear. ^[ref]